A) 30
B) 20
C) 35
D) 25
Correct Answer: A
Solution :
Let the numbers be x, y and z. So, \[x+y=45,\]\[y+z=55,\]\[z+3x=90\] Now, \[y=(45-x)\]and\[z=(55-y)\] \[\therefore \]\[55-(45-x)=10+x\] \[\therefore \]\[10+x+3x=90\]\[\Rightarrow \]\[x=20\] So, third number \[=10+x=10+20=30\]You need to login to perform this action.
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